What is the second derivative of e^xcos x?

Question

What is the second derivative of e^xcos x?

Answer 1

using uv rule, the first derivative is [cosx-sinx]e^x

again using the same, the second derivative is [-sinx-cosx+cosx-sinx]e^x =-2sinx e^x

Answer 2

The first derivative of this expression is:

(e^x)(cos x) - (e^x)(sin x)

To find this derivative on your own, use the product rule. Further explanation can be found at:

http://en.wikipedia.org/wiki/Product_rul...

To find the second derivative of the original expression, take the derivative of the new expression, which is:

-2(e^x)(sin x)

Answer 3

first derivate is:

(e^x·cos(x))'

=(e^x)'·cos(x)+(e^x)·(cos(x))'

=e^x·cos(x)+(e^x)·(-sin(x))

and the second derivate is:

[ e^x·cos(x) + (e^x)·(-sin(x)) ]'

={ (e^x)·cos(x) }' + { (e^x)·(-sin(x)) }'

={ (e^x)'·cos(x) + (e^x)·(cos(x))' } + { (e^x)'·(-sin(x)) + (e^x)·(-sin(x))' }

={ (e^x)·cos(x) + (e^x)·(-sin(x)) } + { (e^x)·(-sin(x)) + (e^x)·(-cos(x)) }

reducing..

={ (e^x)·cos(x) + (e^x)·(-sin(x)) + (e^x)·(-sin(x)) + (e^x)·(-cos(x)) }

={ (e^x)·(cos(x)-sin(x)-sin(x)-cos(x) ) }

={ (e^x)·(-2sin(x)) }

=-2(e^x)(sin(x))

I do not speak English, because of it, only I did only the development

Answer 4

Assuming that this means f'' of (e^x) multiplied by (cos x), then the answer is:

Function cos(x)*(e^x)

f''x =

e^x*

(log(e)^2-1)*

cos(x)-2*

e^x*

log(e)*

sin(x)

The expression has been broken down, because Yahoo! Answers will not display it on a single line.

Hope this helps!